When we speak of black holes and their associated singularity, why is matter drawn into a black hole not condensed into a single point within the singularity?

  • $\begingroup$ @Qmechanic: Ty for edit I was not sure if o could use question in title and just repeat it in the description. $\endgroup$
    – Argus
    May 12, 2012 at 19:36
  • $\begingroup$ What makes think that it is not? $\endgroup$ May 12, 2012 at 19:51
  • $\begingroup$ Well I have been told a few times that is a false assumption. $\endgroup$
    – Argus
    May 12, 2012 at 19:54
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    $\begingroup$ The thing is that you need to make a distinction based on the position of the observer. To a distant observer the infalling mass appears to stack up against the event horizon. This is essentially a time dilation effect. To a infalling observer they and everything before them do rush down to the center where....well, we don't actually know what happens at the singularity. Any way, if the question is about the time dilation it might help to clarify the question some. $\endgroup$ May 12, 2012 at 22:22
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    $\begingroup$ @Argus: If you feel some of your comments are unnecessary, you can remove them yourself. Grenade explosions, however traumatic they are, do not exhibit relativistic effects. $\endgroup$ May 13, 2012 at 4:25

1 Answer 1


The many comments have covered the main points about the question, but I thought it would be worthwhile explaining how the behaviour is calculated. If we solve the Einstein equation for a point mass we get the Schwarzschild metric:

$$ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 + r^2 d\Omega^2$$

All equations look scary to non-nerds, but don't worry too much about the details. The key points are that the equation involves time, $dt$, and the distance from the centre of the black hole, $dr$, and it calculates a quantity called the line element, $ds$.

The time, $t$, and radial distance, $r$, are the physical quantities measured by an observer outside the black hole, i.e. that's you and me. They are exactly what you would think i.e. the time is what we measure with a stopwatch. The radial distance is what you'd get by measuring the circumference of a circle round the black hole and dividing by $2\pi$ (because the circumference of a circle is $2\pi r$).

The line interval, $ds$, is a bit more abstract but for our purposes $ds$ is the time as measured by someone falling into the black hole. This is called the proper time and usually written as $\tau$. You've probably heard that time slows down as you approach the speed of light, and you get a similar effect here (as dmckee menioned in a comment). That means the time measured by someone falling into the black hole, i.e. the proper time $\tau$, is not the same as the time $t$ that we measure when watching the black hole from the outside.

The point of all this stuff is that you can use the metric to calculate how long it takes to fall from some distance, $R$, outside the black hole to the event horizon. First of all let's calculate this for the person falling into the black hole. This means we have to calculate the proper time, $\tau$. You can find this in any book on GR, or by Googling, and the result is:

$$ \Delta \tau = \frac{2M}{3} \left[ \left( \frac{r}{2M} \right)^{3/2} \right] ^R _{2M} $$

Again, don't worry about the detail. A long as we know the mass of the black hole, $M$, and the starting distance, $R$, we just feed these into a calculator and it gives us $\Delta\tau$ which is the time measured by the person falling into the black hole. The time obviously depends on how far away you start and how big the black hole is, but it's just a number of seconds. In fact if we use $r = 0$ in the expression about we can calculate how long it takes to fall through the event horizon and on to the singularity at the black hole.

So, the point to note down at this stage is that the person falling into the black hole reaches the event horizon and inded the singularity in a finite time.

The next step is to calculate the time measured by you and me sitting outside the black hole. This is a bit more involved, but we end up with an expression:

$$ dt = \frac {-(\epsilon + 2M)^{3/2}d\epsilon} {(2M)^{1/2}\epsilon} $$

where $\epsilon$ is the distance from the event horizon i.e. $\epsilon = r - 2M$. Integrating this to find the time to reach the event horizon is a bit messy, but if restrict ourselves to distances very near the event horizon we find:

$$\Delta t \propto ln(\epsilon)$$

but $\epsilon$ is the distance from the event horizon, so it's zero at the event horizon and $ln(0)$ is infinity. That means that the time you and I measure for our astronaut to reach the event horizon, $\Delta t$, is infinity.

And this is why you get the apparently paradoxical result about falling into black holes. The time measured by the person falling in, $\Delta\tau$, is finite but the time measured by people outside the black hole, $\Delta t$, is infinite.

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    $\begingroup$ Okay that was an amazing explination mathematically rigorous and very visually friendly. This makes me think an outside observer can see someone get extremely close to the black hole for simplicity as long as distance from the singularity is not zero which how I see distance from the center of the singularity can never be zero. By this I mean a sphere or a compressed "mass" has a built up layer of particles or smaller than particles touching the very outside edge would mean you are "on" the outside but distance to the internal "center" is still a calculatable value however small in the extreme $\endgroup$
    – Argus
    May 13, 2012 at 15:41
  • $\begingroup$ +1 as I have only a few Rep I can not plus this if someone would not mind bumping it for me? $\endgroup$
    – Argus
    May 13, 2012 at 15:42
  • $\begingroup$ Well I have gained a few rep since breakfast gave the well deserved upvote. $\endgroup$
    – Argus
    May 13, 2012 at 18:18
  • $\begingroup$ ...And this is only about constant (Schwartzschild) metric, not accounting the evaporation of BH. $\endgroup$
    – Anixx
    Nov 10, 2016 at 4:27

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